Question

Really need the answers to these! Please help! State the center point (h, k), the lengths of the major and minor axis, the vertices, the focal points (foci), the eccentricity and graph.
3) 25x^2 + 16y^2 = 400
4) 4x^2 + 9y^2 – 16x + 90y + 97 = 0

Answer 1

first divide everything by 400 to get
\[\frac{x^2}{16}+\frac{y^2}{25}=1\] ok so far?

Answer 2

Yes.

Answer 3

this looks like \(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\) with \(a=4,b=5\) so you know a couple things right away

Answer 4

the center is \((0,0)\)
since the larger number is under the \(y\) terms, it looks like this

Answer 5

and since \(a^2=16\) so \(a=4\) and \(b^2=25\) so \(b=5\) we can label some points

Answer 6

that makes the vertices \((0,5)\) and \((0,-5)\)

Answer 7

so far so good?

Answer 8

Yes

Answer 9

length of major and minor axis now should be obvious. all you need left is the foci, and find that via \(a^2+c^2=b^2\) or \(4^2+c^2=5^2\) making \(c=3\) by either computing or remembering the 3 - 4 - 5 right triangle

Answer 10

since you know the shape of the ellipse, you see that the foci are 3 units up and down from the center
since the center is \((0,0)\) that makes the foci at \((0,3)\) and \((0,-3)\)

Answer 11

what's left?

Answer 12

this make sense so far? the other one is a bit harder but the idea is exactly the same

Answer 13

What would I do since it equals 0?

Answer 14

you mean for the second one?

Answer 15

Yes, sorry

Answer 16

ok this is going to be some work

Answer 17

you have to make
\[4x^2 + 9y^2 – 16x + 90y + 97 = 0\] look like
\[\frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}=1\]

Answer 18

first group the terms, then factor out what you need, then complete the square
it is a drag, but we can work through the steps if you like

Answer 19

I'm working on it

Answer 20

4x(x-4)+9y(y+10)=-97, am I doing this totally wrong?

Answer 21

\[4x^2 + 9y^2 – 16x + 90y + 97 = 0\]\[4x^2-16x+9y^2+90y=-97\] is a start, then
\[4(x^2-4)+9(y^2+10y)=-97\]

Answer 22

now you have to complete the square
\[4(x-2)^2+9(x+5)^2=-97+4\times 2^2+9\times 5^2\]
\[4(x-2)^2+9(x+5)^2=144\] that is the tricky part

Answer 23

let me know if it is not clear where i got those numbers from

Answer 24

once we have that, the rest is more or less routine
divide by 144 to get
\[\frac{(y-2)^2}{36}+\frac{(y+5)^2}{16}=1\]

Answer 25

this tells you the center is \((2,-5)\)

Answer 26

and since \(36>16\) you know it looks like this

Answer 27

ok i had a typo, of course it is
\[\frac{(x-2)^2}{36}+\frac{(y+5)^2}{16}=1\]
the hard part is getting it in that form